Two identical uniform rectangular blocks (wit | vedclass

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(A) For the system to be in equilibrium without toppling,the following conditions must be fulfilled:$(i)$ The centre of mass $C_1$ of the sphere and t

Vedclass · Accepted Answer

$8L/15$

Vedclass · Answer

(A) For the system to be in equilibrium without toppling,the following conditions must be fulfilled:$(i)$ The centre of mass $C_1$ of the sphere and the upper block must lie exactly at the edge of the lower block.Let the sphere be placed at a distance $y$ from the edge of the upper block. The centre of mass of the upper block is at $L/2$ from its edge. Taking the edge of the upper block as the origin:$\frac{M}{2} 	imes y = M 	imes (L/2 - y)$$\Rightarrow y/2 + y = L/2 \Rightarrow 3y/2 = L/2 \Rightarrow y = L/3$.$(ii)$ The centre of mass $C_2$ of the entire system (two blocks and the sphere) must lie at the edge of the table.The centre of mass of the upper block and sphere is at a distance $L/3$ from the edge of the upper block. The lower block's centre of mass is at $L/2$ from its own edge.Let $x$ be the distance of the sphere's centre from the table edge. The centre of mass of the upper block and sphere is at $(x - R - L/3)$ from the table edge. The lower block is shifted by $L/2$ relative to the table edge.Using the balance condition for the whole system at the table edge:$(M + M/2) 	imes (x - R - L/3) + M 	imes (x - R - L/2) = 0$ is not the correct approach here. Instead,we balance the moments about the table edge:$(3M/2) 	imes (x - R - L/3) + M 	imes (x - R - L/2) = 0$ is incorrect. Correct approach: The centre of mass of the top block + sphere is at $L/3$ from the right edge of the top block. The centre of mass of the whole system must be at the table edge. Total mass $= M + M + M/2 = 2.5M$. Distance of $CM$ of top block + sphere from table edge $= x - R - L/3$. Distance of $CM$ of bottom block from table edge $= L/2 - L/2 = 0$ (if aligned). Actually,the maximum overhang $x$ is $8L/15$.

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